# State De Moivre's theorem and prove it for all integer values?

May 3, 2018

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

Proof by induction

#### Explanation:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

Prove by induction:

Basis case: $n = 1$

$\implies {\left(\cos \theta + i \sin \theta\right)}^{1} = \cos \theta + i \sin \theta$

Hence basis case holds

Asume true for $n = k$

${\left(\cos \theta + i \sin \theta\right)}^{k} = \cos k \theta + i \sin k \theta$

Showing holds for $n = k + 1$

${\left(\cos \theta + i \sin \theta\right)}^{k + 1} = \left(\cos k \theta + i \sin k \theta\right) \left(\cos \theta + i \sin \theta\right)$

${\left(\cos \theta + i \sin \theta\right)}^{k + 1} =$

$= \cos \theta \cos k \theta + i \sin \theta \cos k \theta + i \sin k \theta \cos \theta - \sin k \theta \sin \theta$

$= \cos \theta \cos k \theta - \sin k \theta \sin \theta + i \left(\sin \theta \cos k \theta + \sin k \theta \cos \theta\right)$

$\implies = \cos \left(\left(k + 1\right) \theta\right) + i \sin \left(\left(k + 1\right) \theta\right)$

Hence it holds true for $k + 1$

It holds for basis case $n = 1$ and $n = k + 1$ for all $k$ that holds, hence holds for $n = 1$, $n = 2$ , $n = 3$ etc...

$\implies \text{holds} \forall n \in \mathbb{Z}$

May 3, 2018

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) \forall n \in \mathbb{R}$

#### Explanation:

De Moivre's Theorem, states that:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) \forall n \in \mathbb{R}$

We seek to validate this result for $n \in \mathbb{N}$, which we can prove using Mathematical Induction:

Induction Proof - Hypothesis

We seek to prove that:



${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) \forall n \in \mathbb{N} \setminus \setminus \setminus \setminus$ ..... [A]

So let us test this assertion, [A], using Mathematical Induction:

Induction Proof - Base case:

Consider, the special case $n = 1$ in which case we have the trivial result:

${\left(\cos \theta + i \sin \theta\right)}^{1} = \cos \left(1 \theta\right) + i \sin \left(1 \theta\right)$

So the given result is true when $n = 1$.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when $n = m$, for some $m \in \mathbb{N} , m > 1$, in which case for this particular value of $m$ we have:

${\left(\cos \theta + i \sin \theta\right)}^{m} = \cos \left(m \theta\right) + i \sin \left(m \theta\right) \setminus \setminus \setminus$ ..... [B]

Then, consider the expression:

$E = {\left(\cos \theta + i \sin \theta\right)}^{m + 1}$

$\setminus \setminus \setminus = {\left(\cos \theta + i \sin \theta\right)}^{m} \left(\cos \theta + i \sin \theta\right)$

$\setminus \setminus \setminus = \left(\cos \left(m \theta\right) + i \sin \left(m \theta\right)\right) \left(\cos \theta + i \sin \theta\right) \setminus \setminus$ (using [B])

$\setminus \setminus \setminus = \cos \left(m \theta\right) \cos \theta + i \sin \left(m \theta\right) \cos \theta + \cos \left(m \theta\right) i \sin \theta + i \sin \left(m \theta\right) i \sin \theta$

$\setminus \setminus \setminus = \left\{\cos \left(m \theta\right) \cos \theta - \sin \left(m \theta\right) \sin \theta\right\} + i \left\{\sin \left(m \theta\right) \cos \theta + i \cos \left(m \theta\right) \sin \theta\right\}$

$\setminus \setminus \setminus = \cos \left(m + 1\right) \theta + i \sin \left(m + 1\right) \theta$

Which is the given result [A] with $n = m + 1$

Induction Proof - Summary

So, we have shown that if the given result [A] is true for $n = m$, then it is also true for $n = m + 1$. But we initially showed that the given result was true for $n = 1$ so it must also be true for $n = 2 , n = 3 , n = 4 , \ldots$ and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for $n \in \mathbb{N}$

Hence we have:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) \forall n \in \mathbb{N} \setminus \setminus \setminus \setminus$ QED

May 3, 2018

See below.

#### Explanation:

First let us prove $r \left[\cos \theta + j \sin \theta\right]$=$r {e}^{j \theta}$.

Let $z = \cos \theta + j \sin \theta \ldots \ldots \ldots . .$[1], therefore, dz/[d theta$=$-[sintheta-jcostheta]

=${j}^{2} \sin \theta + j \cos \theta$ = $j \left[\cos \theta + j \sin \theta\right]$[ sincej^2=-1]

But $j \left[\cos \theta + j \sin \theta\right]$=$z$ [ from .....$\left[1\right]$, so dz/[d theta=$j z$.........$\left[2\right]$
Inverting both sides of .....$\left[2\right]$, d theta/[dz=1/[jz# and integrating w.r.t $z$

$\theta = \frac{1}{j} \int \frac{1}{z} \mathrm{dz}$ = $\frac{1}{j} \ln z + c$, but from .....$\left[1\right]$ when $\theta = 0 , \cos \theta = 1 , \sin \theta = 0 , i . e . , z = 1$

Therefore, $\theta = \frac{1}{j} \ln \left[1\right] + c$, but $\ln \left[1\right] = 0$ to any base, therefore $c = 0$.

So we have, $\theta = \frac{1}{j} \ln z$, i.e, $j \theta = \ln x$, and so ${e}^{j} \theta = z$ [ theory of logs], therefore from ...$\left[1\right]$, $r \left[\cos \theta + j \sin \theta\right] = r {e}^{j} \theta$.......$\left[3\right]$

[ Proof of De Moivres theorem, for any value of $n$]

From .....$\left[3\right]$, $\cos \theta + j \sin \theta = {e}^{j} \theta$ [ omitting $r$].......$\left[4\right]$

Raising both sides of ....$\left[3\right]$ to the power $n$, ${\left[\cos \theta + j \sin \theta\right]}^{n}$=${e}^{n j \theta}$

=${e}^{j \left[n \theta\right]}$ , but ${e}^{j \left[n \theta\right]}$= $\cos n \theta + j \sin n \theta$ , from .......$\left[4\right]$, replacing $\theta$ by $n \theta$,

Therefore , ${\left[\cos \theta + j \sin \theta\right]}^{n}$= $\cos n \theta + j \sin n \theta$.